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2 years ago

A truck and a scooter moving with the same velocity colide. Why does the scooter fly off?

1 answer:

6 0

We know, Momentum is the product of mass and velocity. As velocity is equal, momentum of Truck will be greater as mass of Truck is greater than the scooter. As momentum is greater from truck side, it will transfer the same amount to the scooter, so it flies away

Hope this helps!

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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

2 years ago

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What type of weather would be in the middle of the picture, where the barometric pressure is 1000 millibars?

UkoKoshka [18]

It would be D. Hope that helped :)

4 0

2 years ago

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

3 years ago

S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(

Ket [755]

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.

V = kq/r

Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.

To learn more about electric potential refer to:

brainly.com/question/12645463

#SPJ4

3 0

1 year ago

What s physics?

Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

7 0

2 years ago

Read 2 more answers